The integration of ?int((x^2-1)/(x^3-3x)) dx

int((x^2-1)/(x^3-3x)) dx

2 Answers
Apr 13, 2018

The answer is =1/3ln(|x^3-3x|)+C

Explanation:

Perform this integral by substitution

Let u=x^3-3x

du=3x^2-3=3(x^2-1)dx

Therefore, the integral is

int((x^2-1)dx)/(x^3-3x)=1/3int(du)*1/u

=1/3lnu

=1/3ln(|x^3-3x|)+C

Apr 13, 2018

1/3ln|(x^3-3x)|+C, or, ln|root(3)(x^3-3x)|+C.

Explanation:

Suppose that, I=int(x^2-1)/(x^3-3x)dx.

Observe that, d/dx(x^3-3x)=3x^2-3=3(x^2-1).

Knowing that, int(f'(x))/f(x)dx=ln|f(x)|+c, we have,

I=1/3int(3(x^2-1))/(x^3-3x)dx.

rArr I=1/3ln|(x^3-3x)|+C, or, ln|root(3)(x^3-3x)|+C.